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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
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-\begin_preamble
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-\index Index
-\shortcut idx
-\color #008000
-\end_index
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc1[00]
-\end_layout
-
-\end_inset
-
-What is the smallest positive rational number?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-There isn't.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc2[00]
-\end_layout
-
-\end_inset
-
-Is 
-\begin_inset Formula $1+0.239999999\dots$
-\end_inset
-
- a decimal expansion?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Yes,
- as long as it means a number 
-\begin_inset Formula $x$
-\end_inset
-
- with 
-\begin_inset Formula $1.239999999\leq x<1.24$
-\end_inset
-
-,
- that is,
- as long as it doesn't end with an infinite number of nines.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc3[02]
-\end_layout
-
-\end_inset
-
-What is 
-\begin_inset Formula $(-3)^{-3}$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $(-3)^{-3}=\frac{1}{(-3)^{3}}=\frac{1}{-27}=-\frac{1}{27}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc4[05]
-\end_layout
-
-\end_inset
-
-What is 
-\begin_inset Formula $(0.125)^{-2/3}$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $(0.125)^{-2/3}=(\frac{1}{8})^{-2/3}=\sqrt[3]{(\frac{1}{8})^{-2}}=\sqrt[3]{8^{2}}=\sqrt[3]{64}=4$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc5[05]
-\end_layout
-
-\end_inset
-
-We defined real numbers in terms of a decimal expansion.
- Discuss how we could have defined them in terms of a binary expansion instead,
- and give a definition to replace Eq.
- (2).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-A decimal expansion would have the form 
-\begin_inset Formula $x=n+0.b_{1}b_{2}b_{3}\dots$
-\end_inset
-
-,
- where each 
-\begin_inset Formula $b_{i}$
-\end_inset
-
- would be a binary digit,
- 0 or 1,
- and this would mean that
-\begin_inset Formula 
-\[
-n+\frac{b_{1}}{2}+\frac{b_{2}}{4}+\dots+\frac{b_{k}}{2^{k}}\leq x<n+\frac{b_{1}}{2}+\frac{b_{2}}{4}+\dots+\frac{b_{k}}{2^{k}}+\frac{1}{10^{k}}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc6[10]
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $x=m+0.d_{1}d_{2}\dots$
-\end_inset
-
- and 
-\begin_inset Formula $y=n+0.e_{1}e_{2}\dots$
-\end_inset
-
- be real numbers,
- Give a rule for determining whether 
-\begin_inset Formula $x=y$
-\end_inset
-
-,
- 
-\begin_inset Formula $x<y$
-\end_inset
-
-,
- or 
-\begin_inset Formula $x>y$
-\end_inset
-
-,
- based on the decimal representation.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We say that 
-\begin_inset Formula $x=y$
-\end_inset
-
- if 
-\begin_inset Formula $m=n$
-\end_inset
-
- and 
-\begin_inset Formula $d_{k}=e_{k}$
-\end_inset
-
- for all 
-\begin_inset Formula $k$
-\end_inset
-
-,
- and that 
-\begin_inset Formula $x<y$
-\end_inset
-
- if 
-\begin_inset Formula $m<n$
-\end_inset
-
- or if 
-\begin_inset Formula $m=n$
-\end_inset
-
- and there's a 
-\begin_inset Formula $k$
-\end_inset
-
- such that 
-\begin_inset Formula $d_{j}=e_{j}$
-\end_inset
-
- for 
-\begin_inset Formula $1\leq j<k$
-\end_inset
-
- and 
-\begin_inset Formula $d_{k}<e_{k}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc11[10]
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $b=10$
-\end_inset
-
- and 
-\begin_inset Formula $x\approx\log_{10}2$
-\end_inset
-
-,
- to how many decimal places of accuracy will we need to know the value of 
-\begin_inset Formula $x$
-\end_inset
-
- in order to determine the first three decimal places of the decimal expansion of 
-\begin_inset Formula $b^{x}$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Potentially infinite.
- Because 
-\begin_inset Formula $10^{\log_{10}2}=2$
-\end_inset
-
-,
- 
-\begin_inset Formula $10^{x}=2.0\dots$
-\end_inset
-
- if 
-\begin_inset Formula $x\geq\log_{10}2$
-\end_inset
-
- or 
-\begin_inset Formula $10^{x}=1.9\dots$
-\end_inset
-
- if 
-\begin_inset Formula $x\leq\log_{10}2$
-\end_inset
-
-,
- but since 
-\begin_inset Formula $\log_{10}2$
-\end_inset
-
- is irrational,
- its decimal expansion is infinite and,
- in the worst case that 
-\begin_inset Formula $x=\log_{10}2$
-\end_inset
-
-,
- we can't determine by an algorithm that 
-\begin_inset Formula $x\geq\log_{10}2$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc12[02]
-\end_layout
-
-\end_inset
-
-Explain why Eq.
- (10) follows from Eqs.
- (8).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Because logarithms are strictly increasing,
- continuous functions,
- and we have that 
-\begin_inset Formula $10^{0.30102999}<2<10^{0.30103000}$
-\end_inset
-
-,
- we can take logarithms on the expression to get 
-\begin_inset Formula $0.30102999<\log_{10}2<0.30103000$
-\end_inset
-
-,
- and so 
-\begin_inset Formula $\log_{10}2=0.30102999$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc13[M23]
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-Given that 
-\begin_inset Formula $x$
-\end_inset
-
- is a positive real number and 
-\begin_inset Formula $n$
-\end_inset
-
- is a positive integer,
- prove the inequality 
-\begin_inset Formula $\sqrt[n]{1+x}-1\leq x/n$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-Use this fact to justify the remarks following (7).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-Let 
-\begin_inset Formula $f(x):=\sqrt[n]{1+x}-1-\frac{x}{n}$
-\end_inset
-
-,
- we have to prove that,
- if 
-\begin_inset Formula $n$
-\end_inset
-
- is a positive integer and 
-\begin_inset Formula $x>0$
-\end_inset
-
-,
- then 
-\begin_inset Formula $f(x)\leq0$
-\end_inset
-
-.
- We have 
-\begin_inset Formula 
-\[
-f'(x)=\frac{1}{n\sqrt[n]{1+x}}-\frac{1}{n}=\frac{1}{n}\left(\frac{1}{\sqrt[n]{1+x}-1}\right),
-\]
-
-\end_inset
-
-but since
-\begin_inset Formula $\sqrt[n]{1+x}>1$
-\end_inset
-
- for 
-\begin_inset Formula $x>0$
-\end_inset
-
-,
- we have 
-\begin_inset Formula $f'(x)<0$
-\end_inset
-
- and 
-\begin_inset Formula $f$
-\end_inset
-
- is strictly decreasing on 
-\begin_inset Formula $(0,+\infty)$
-\end_inset
-
-.
- Since 
-\begin_inset Formula $f(0)=0$
-\end_inset
-
-,
- this proves the result.
-\end_layout
-
-\begin_layout Enumerate
-It's clear that 
-\begin_inset Formula $b^{n+d_{1}/10+\dots+d_{k}/10^{k}}\leq b^{n+1}$
-\end_inset
-
-,
- and then,
- taken 
-\begin_inset Formula $x=b-1>0$
-\end_inset
-
- and 
-\begin_inset Formula $n=10^{k}$
-\end_inset
-
-,
- we have 
-\begin_inset Formula $\sqrt[10^{k}]{1+b-1}=b^{1/10^{k}}\leq(b-1)/10^{k}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc15[10]
-\end_layout
-
-\end_inset
-
-Prove or disprove:
-\begin_inset Formula 
-\begin{align*}
-\log_{b}x/y & =\log_{b}x-\log_{b}y, & \text{if }x,y & >0.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $\alpha:=\log_{b}x$
-\end_inset
-
- and 
-\begin_inset Formula $\beta:=\log_{b}y$
-\end_inset
-
-,
- we have
-\begin_inset Formula 
-\[
-\frac{b^{\alpha}}{b^{\beta}}=b^{\alpha}b^{-\beta}=b^{\alpha-\beta},
-\]
-
-\end_inset
-
-and taking logarithms,
- we get 
-\begin_inset Formula $\log_{b}\frac{b^{\alpha}}{b^{\beta}}=\log_{b}\frac{x}{y}=\log_{b}b^{\alpha-\beta}=\alpha-\beta=\log_{b}x-\log_{b}y.$
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc16[00]
-\end_layout
-
-\end_inset
-
-How can 
-\begin_inset Formula $\log_{10}x$
-\end_inset
-
- be expressed in terms of 
-\begin_inset Formula $\ln x$
-\end_inset
-
- and 
-\begin_inset Formula $\ln10$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $\log_{10}x=\frac{\log_{e}x}{\log_{e}10}=\frac{\ln x}{\ln10}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc17[05]
-\end_layout
-
-\end_inset
-
-What is 
-\begin_inset Formula $\lg32$
-\end_inset
-
-?
- 
-\begin_inset Formula $\log_{\pi}\pi$
-\end_inset
-
-?
- 
-\begin_inset Formula $\ln e$
-\end_inset
-
-?
- 
-\begin_inset Formula $\log_{b}1$
-\end_inset
-
-?
- 
-\begin_inset Formula $\log_{b}(-1)$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $\lg32=5$
-\end_inset
-
-,
- 
-\begin_inset Formula $\log_{\pi}\pi=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $\ln e=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $\log_{b}1=0$
-\end_inset
-
-.
- As for 
-\begin_inset Formula $\log_{b}(-1)$
-\end_inset
-
-,
- it would be an 
-\begin_inset Formula $x$
-\end_inset
-
- such that 
-\begin_inset Formula $b^{x}=-1$
-\end_inset
-
-,
- which is not a real number most of the time.
- As a complex number,
- it would be 
-\begin_inset Formula $\log_{b}(-1)=\frac{\ln(-1)}{\ln b}=\frac{i\pi}{\ln b}$
-\end_inset
-
-,
- since 
-\begin_inset Formula $e^{i\pi}=-1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc18[10]
-\end_layout
-
-\end_inset
-
-Prove or disprove:
- 
-\begin_inset Formula $\log_{8}x=\frac{1}{2}\lg x$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $\log_{8}x=\frac{\lg x}{\lg8}$
-\end_inset
-
-,
- but 
-\begin_inset Formula $\lg8=3\neq2$
-\end_inset
-
-,
- so this is false in the general case (this is only true when 
-\begin_inset Formula $x=1$
-\end_inset
-
-).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc19[20]
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $n$
-\end_inset
-
- is an integer whose decimal representation is 14 digits long,
- will the value of 
-\begin_inset Formula $n$
-\end_inset
-
- fit in a computer word with a capacity of 47 bits and a sign bit?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The number of possibilities for a number with up to 14 digits is 
-\begin_inset Formula $10^{14}$
-\end_inset
-
-,
- and the number for 47 bits is 
-\begin_inset Formula $2^{47}\approx1.3\cdot2^{14}$
-\end_inset
-
-,
- so it would fit.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc20[10]
-\end_layout
-
-\end_inset
-
-Is there any simple relation between 
-\begin_inset Formula $\log_{10}2$
-\end_inset
-
- and 
-\begin_inset Formula $\log_{2}10$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $\log_{10}2=\frac{\log_{2}2}{\log_{2}10}=\frac{1}{\log_{2}10}$
-\end_inset
-
-,
- so 
-\begin_inset Formula $\log_{10}2\log_{2}10=1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc22[20]
-\end_layout
-
-\end_inset
-
-(R.
- W.
- Hamming.) Prove that
-\begin_inset Formula 
-\[
-\lg x\approx\ln x+\log_{10}x,
-\]
-
-\end_inset
-
-with less than 
-\begin_inset Formula $\unit[1]{\%}$
-\end_inset
-
- error!
- (Thus a table of natural logarithms and of common logarithms can be used to get approximate values of binary logarithms as well.)
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\[
-\ln x+\log_{10}x=\frac{\lg x}{\lg e}+\frac{\lg x}{\lg10}=\lg x\left(\frac{1}{\lg e}+\frac{1}{\lg10}\right)\cong0.994\lg x\approx\lg x,
-\]
-
-\end_inset
-
-and this gives us about 
-\begin_inset Formula $\unit[0.6]{\%}$
-\end_inset
-
- error.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc27[M25]
-\end_layout
-
-\end_inset
-
-Consider the method for calculating 
-\begin_inset Formula $\log_{10}x$
-\end_inset
-
- discussed in the text.
- Let 
-\begin_inset Formula $x'_{k}$
-\end_inset
-
- denote the computed approximation to 
-\begin_inset Formula $x_{k}$
-\end_inset
-
-,
- determined as follows:
- 
-\begin_inset Formula $x(1-\delta)\leq10^{n}x'_{0}\leq x(1+\epsilon)$
-\end_inset
-
-;
- and in the determination of 
-\begin_inset Formula $x'_{k}$
-\end_inset
-
- by Eqs.
- (18),
- the quantity 
-\begin_inset Formula $y_{k}$
-\end_inset
-
- is used in place of 
-\begin_inset Formula $(x'_{k-1})^{2}$
-\end_inset
-
-,
- where 
-\begin_inset Formula $(x'_{k-1})^{2}(1-\delta)\leq y_{k}\leq(x'_{k-1})^{2}(1+\epsilon)$
-\end_inset
-
- and 
-\begin_inset Formula $1\leq y_{k}<100$
-\end_inset
-
-.
- Here 
-\begin_inset Formula $\delta$
-\end_inset
-
- and 
-\begin_inset Formula $\epsilon$
-\end_inset
-
- are small constants that reflect the upper and lower errors due to rounding or truncation.
- If 
-\begin_inset Formula $\log^{\prime}x$
-\end_inset
-
- denotes the result of the calculations,
- show that after 
-\begin_inset Formula $k$
-\end_inset
-
- steps we have
-\begin_inset Formula 
-\[
-\log_{10}x+2\log_{10}(1-\delta)-1/2^{k}<\log^{\prime}x\leq\log_{10}x+2\log_{10}(1+\epsilon).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We first prove by induction that
-\begin_inset Formula 
-\[
-x^{2^{k}}(1-\delta)^{2^{k+1}-1}\leq10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k}\leq x^{2^{k}}(1+\epsilon)^{2^{k+1}-1}.
-\]
-
-\end_inset
-
-For 
-\begin_inset Formula $k=0$
-\end_inset
-
-,
- we have 
-\begin_inset Formula $x(1-\delta)\leq10^{n}x'_{0}\leq x(1+\epsilon)$
-\end_inset
-
-,
- which is given.
- If this holds up to a certain 
-\begin_inset Formula $k$
-\end_inset
-
-,
- then if 
-\begin_inset Formula $(x'_{k})^{2}<10$
-\end_inset
-
-,
- we have 
-\begin_inset Formula $b_{k+1}=0$
-\end_inset
-
-,
- 
-\begin_inset Formula $(1-\delta)(x'_{k})^{2}\leq x'_{k+1}\leq(1+\epsilon)(x'_{k})^{2}$
-\end_inset
-
- and
-\begin_inset Formula 
-\begin{eqnarray*}
-x^{2^{k+1}}(1-\delta)^{2^{k+2}-1} & = & (x^{2^{k+1}}(1-\delta)^{2^{k+1}-1})^{2}(1-\delta)\\
- & \leq & (10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k})^{2}(1-\delta)\\
- & \leq & 10^{2^{k+1}(n+b_{1}/2+\dots+b_{k}/2^{k}+b_{k+1}/2^{k+1})}x'_{k+1}\\
- & \leq & (10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k})^{2}(1+\epsilon)\\
- & \leq & (x^{2^{k}}(1+\epsilon)^{2^{k+1}-1})^{2}(1+\epsilon)=x^{2^{k+1}}(1+\epsilon)^{2^{k+2}-1}.
-\end{eqnarray*}
-
-\end_inset
-
-If 
-\begin_inset Formula $(x'_{k})^{2}\geq10$
-\end_inset
-
-,
- we have 
-\begin_inset Formula $b_{k+1}=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $(1-\delta)(x'_{k})^{2}\leq10x'_{k+1}\leq(1+\epsilon)(x'_{k})^{2}$
-\end_inset
-
- and
-\begin_inset Formula 
-\begin{eqnarray*}
-x^{2^{k+1}}(1-\delta)^{2^{k+2}-1} & \leq & (10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k})^{2}(1-\delta)\\
- & \leq & 10^{2^{k+1}(n+b_{1}/2+\dots+b_{k}/2^{k})}10x'_{k+1}\\
- & = & 10^{2^{k+1}(n+b_{1}/2+\dots+b_{k}/2^{k}+b_{k+1}/2^{k+1})}x'_{k+1}\\
- & \leq & (10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k})^{2}(1+\epsilon)\\
- & \leq & x^{2^{k+1}}(1+\epsilon)^{2^{k+2}-1}.
-\end{eqnarray*}
-
-\end_inset
-
-Then,
- by taking logarithms on the expression,
- we get
-\begin_inset Formula 
-\[
-2^{k}\log x+(2^{k+1}-1)\log(1-\delta)\leq2^{k}\log^{\prime}x+\log x'_{k}\leq2^{k}\log x+(2^{k+1}-1)\log(1+\epsilon).
-\]
-
-\end_inset
-
-Finally,
- subtracting 
-\begin_inset Formula $\log x'_{k}$
-\end_inset
-
- from all sides,
- using that 
-\begin_inset Formula $-1<-\log(1-\delta)-\log x'_{k}=-\log(x'_{k}(1-\delta))$
-\end_inset
-
-,
- because 
-\begin_inset Formula $x'_{k}(1-\delta)<10$
-\end_inset
-
-,
- using that 
-\begin_inset Formula $-\log(1+\epsilon)-\log x'_{k}=-\log(x'_{k}(1+\epsilon))\leq0$
-\end_inset
-
-,
- because 
-\begin_inset Formula $x'_{k}(1+\epsilon)\geq1$
-\end_inset
-
-,
- and dividing everything by 
-\begin_inset Formula $2^{k}$
-\end_inset
-
-,
- we get
-\begin_inset Formula 
-\[
-\log x+2\log(1-\delta)-\frac{1}{2^{k}}<\log^{\prime}x\leq\log x+2\log(1+\epsilon),
-\]
-
-\end_inset
-
-which is precisely what we wanted to prove.
-\end_layout
-
-\end_body
-\end_document